@copper.hat No. It is the language spoken by swedish chef.

Bork bork coborkism bork.

:-). When I arrived in the USA in 1983, I often had to repeat myself multiple times in order to be understood.

huh?

:)

I'm studying first course in algebra and although it has been a month, I just see that the behaviour of the multiplication xN*yN in a quotient group is the same as if we would use the corresponding homomorphism instead of N with whatever it applies to the element

and the quotient group seem to "factor" the homomorphism similar to how we use to factor .e.g 6=2*3 and ignore the multiplication by 1

does it make sense or I'm just confusing

@X4J Any time you have a normal subgroup of your normal subgroup, you may factor through the first one as the natural map you define becomes well-defined. That's used a lot in advanced topics.

I'm confusing myself though, so take this with a grain of salt

I mean $G/K \to G/N$ is well defined whenever $K \leqslant N \leqslant G$ all normal

This is because of well-definedness of $g + K \mapsto \pi(g)$ where $\pi : G \to G/N$ is your quotient natural projection

This is true since $\pi$ takes the $K$ to zero (being a subset of the kernel $N$ of $\pi$)

Use that to prove that if $h + K = g + K$ (two reps of the same equivalence class) then $f(h) = f(g)$ where $f : G/K \to G/N$ is the map we're trying to prove well-defined.

ANY TIME you have a quotient taken, you always have to prove that maps FROM IT are well-defined.

Project I'm working on

Thing is I thought I understood it but now it clicks deeper

So essentially, the mnemonic is this: Given a group (ring, module, etc) homomorphism $g : G \to H$, if you have a (normal) subgroup of the kernel of $g$, then you may factor $g$ through it! Forming $g = f \circ h$, where $f: G \to G/K$ and $h : G/K \to H$.

I mean $h \circ f$

All that needs to be proven by you though before you become confident

It's just namely well-definedness out of the quotient that can fail

Well-definedness simply means that it's actually a function and agrees on any choice of representative

It's used all the time though, say in HA (homological algebra)

Well defined ie that it preserves the structure in the quotient group?

No that's the second thing. Once you have well-definedness proven, you have to prove that your function is actually a group hom

I think I know what your question was about though.

You can work equivalently in $G$ or in $G/N$ in order to work in $G/N$ but after each computation in $G$ you need to "modulo by $N$" in order to get your result in $G/N$.

And the importance of studying it is to be able to focus on certain properties of the group in a simpler form?

It's like how modular arithmetic works because that's a special case of this general phenomenon

Not sure, quotient groups are broadly applicable to every area of math

They sometimes use it in some proofs as a way to do induction because $G/N$ is usually strictly smaller in cardinality than $G$

Yes but here it captures the homomorphisms

Not sure what you mean by that, pls explain

Oh sorry I confused

No worries 😵

I mean why would one want to abstract the notion of homomorphism

In practice

Homomorphism just means structure preserving map in algebra

In category theory they've abstracted this a little

@DanielDonnelly yeah thats exactly it

Some things can be proven without examining the elements but the proofs seem very alien to me and hard to find

They call that pure arrow-based math

So you have the category $\textbf{Ab}$ of abelian groups. This forms an abelian category. In a general abelian category there is no notion of elements other than what you can model with arrows ("generalized elements"). However because of Freyd-Mitchell embedding theorem you may do diagram chases in some category of $R$-modules (essentially vector spaces without invertibility of scalars) and that suffices to do it in your original abelian category (very general / abstract)

So abstraction is good but not at the expense of not being able to actually do the math. Which is why most diagram chase proofs let you take elements at the nodes (abelian groups)

Btw arrows = homomorphisms here

I apparently have to give a talk on some math topic sometime next month, mostly for undergrad audience (from 1st to 3rd year, mostly). I was thinking I will do some reading on AC, Zorns, cardinal / ordinal stuff and speak on that. thing is that shit is daunting even for me, the dude whos supposed to give the talk xD

do you want to talk about any theorem?

In this proof of the statement: "no natural number is equivalent to it's proper subset". I'm stuck at the point where $n \in E$. Can someone explain why n is equivalent to $E-{n}$?

anyone?

@ShyamTripathi they assumed that $n^+$ is equivalent to $E$. As $n^+=n\cup \{n\}$ so $n\cup \{n\}$ is equivalent to $E$, implying $n$ is equivalent to $E\setminus \{n\}$

Oh! Thanks! I did not know if that is a valid arithmetic on equivalent sets. So if I have $A \cup B$ equivalent to C then $A$ is equivalent to $C-B$?

@SoumikMukherjee

I'll try proving this if it is true.

@ShyamTripathi no

take A={1,2}, B={2,3}, C={1,2,3}

Ok.

the element {n} is in both of the set, so removing it won't cause a problem

@SoumikMukherjee Oh! Thanks! This is what I was looking for.

np

@SoumikMukherjee yeah ideally

I think i can talk abt ac, zorns and their equivalence. speak abt vector spaces n stuff, but perhaps more interesting may be comparision and arithmetic of cardinal, which I should familiarise myself with first

Is there any $a\in (0,1)-{\frac{1}{2}\}$ that we have a closed form for $\Gamma(a)$?

@nickbros123 cool

@pie you mean \

But I have to do a lot of reading myself

yeah

use \setminus

@SoumikMukherjee TBH the $-$ is better

project: become Asaf Karagila in 1 month

@nickbros123 well...

@pie I like \

anyway I can't find any such $a$ in en.wikipedia.org/wiki/Particular_values_of_the_gamma_function

this is not just a knowledge problem, its a pedagogy problem as well, I think. I dont think I can just speak on theorems and hope to keep the attention, think there should be some structure to it. maybe some kind of payoff at the near the end or something

maybe it doesn't exist, I am curious about this, do you think that I should make an MSE question? do you think such a question will do well on MSE?

Hi

Hi

how are you ?

Fine, thank for asking. What about you?

@nickbros123 mini Asaf Karagila. Ordinals, cardinals, that's just the beginning really

set theorists do some much crazier stuff

@Jakobian fair

even I encounter crazier stuff and I'm not a set theorist :P

@SineoftheTime im fine, this month I have to take 2 exams, I hope well

if I pass then they will think about what to study

I (regretfully) haven't dedicated much time to the technicalities of set-theory or model-theory. My personal way of dealing with set-theoretic issues is to take a strongly inaccessible cardinal $\kappa$, and take the associated Von Neumann universe $\mathfrak{U}:=V_\kappa$, and call elements of this $\mathfrak{U}$-universe, subset of it $\mathfrak{U}$-classes, and any ZFC set that is not in bijection with such a $\mathfrak{U}$-class is called a conglomerate. What I wonder about is whether [cont]

Consider the definition of total variation of a signed measure $\mu$ (which is map from $\mathcal{A}\to\mathbb R$): $$|\mu|(A)=\sup\left\{\sum_{n\in\mathbb N}|\mu(A_n)|:A=\bigcup_{n\in\mathbb N}A_n,A_n\in\mathcal{A}, A_n\text{ disjoint}\right\}.$$

To show that $|\mu|$ is a measure, the author says $|\mu|(\varnothing)=0$ is obvious. Ok. Then, let $(B_i)_{i\in\mathbb N}$ be a sequence of disjoint $\mathcal{A}$-measurable sets (we want to show countable additivity of $|\mu|$). Set $B=\bigcup_{i\in\mathbb N}B_i$ The author says we can find $t_i\in [0,|\mu|(B_i))$ and write $B_i=\bigcup_{n\in\mathbb N}A_{n,i}$ for disjoint $(A_{n,i})_{n,i\in\mathbb N}$.

Then through the given facts, he derives that $|\mu|(B)\geq\sum_{i\in\mathbb N}t_i$ and says we can choose $t_i$ arbitrarily close to $|\mu|(B_i)$ so that the previous inequality reads $|\mu|(B)\geq\sum_{i\in\mathbb N}|\mu|(B_i)$. I figure, we choose $t_i=|\mu|(B_i)-\epsilon2^{-i}$ for some $\epsilon>0$ small, but what if $|\mu|(B_i)$ decreases faster than $\epsilon2^{-i}$ so that $t_i$ eventually becomes negative?

[cont] one has the axiom of specification in a $\mathfrak{U}$-class. It seems you can take a larger Von Neumann universe $\mathfrak{V}$, and in this larger universe a $\mathfrak{U}$-class will now be a $\mathfrak{V}$-set, and this being a model of ZFC, I have the axiom of specification there, and can pick out sub-$\mathfrak{V}$-sets in a set-builder fashion. But such a sub-$\mathfrak{V}$-set is certainly still a subset of our $\mathfrak{U}$-class, so must be a $\mathfrak{U}$-class? [cont]

[cont] So I do have the axiom of specification for $\mathfrak{U}$-classes?

But you seem correct, if $A$ is a $\mathfrak{U}$-class and $\varphi$ is a formula in ZFC with $x$ a free variable, then there is set $S$ such that $x\in S\iff \varphi(x)\land x\in A$, so that $S\subseteq A\subseteq\mathfrak{U}$ and so $S\subseteq\mathfrak{U}$ which is by definition what it means to be a $\mathfrak{U}$-class

Well we probably want $\varphi$ to be a "formula in $\mathfrak{U}$" however that would be defined, and then trivially extend it to a formula in ZFC

@psie does it matter that $t_i$ becomes negative, other than the author wanted for $t_i$ to be in $[0, |\mu|(B_i))$?

@Jakobian well, I would say no, it doesn't matter actually. It's weird that $t_i\in [0,|\mu|(B_i))$ is stipulated, not really sure what its purpose.

I think so too

👍

Let X be a compact (sequentially) metric space. If $f: X \rightarrow X$ satisfies d(f(x), f(y)) < d(x,y) for each $x \noteq y$ in the space X then f has a fixed-point.

Can someone guide me thro solving it? I have no idea where to go right now

ah, this is a classic

you can prove it using that continuous real-valued function on a compact space admits a supremum iirc

Yes I tried to do that but I struggle to figure the composition that takes X to R

I guess it should involve the metric and some guessed point?

@X4J hint d(x,f(x))

don't you need the completenss hypothesis?

oh, you have compactness

It's not true for complete metric spaces

do you know Banach's fixed point theorem

Yeah I do

you can reduce this to that

but that theorem is more intuitive to me and I've met this notion when I questioned here math.stackexchange.com/questions/4763306/…

however this question seems different in the sense that the distance does not get smaller geometrically

oh it should imply this geometric behavior too?

@SoumikMukherjee yeah. Compactness is important here

it can be annoying to determine if theorems of this type are true sometimes. Thankfully, in this case this is easy

Since it's false if we change the assumption of being compact with complete metric space

you can check out a (certainly not exhaustive) book by Dugundji and Granas under the faithful title "Fixed point theory"

there is no 'fixed point theory' as far as I know, but there are various ways in which people study fixed points, be it in order theory, or in functional analysis etc.

Let $M^n$ be a homology $n$-manifold. Does it follows $H_i(M^n)= \mathbb{Z}$? for $I=n$ and $0$ $I\neq n$?

I have question, how do people hear study math? what general advice would u give from experience? (would love to hear answers from diff people) :)

@Pizza linear algebra and physics?

@monoidaltransform Torus is a manifold (and therefore a homology manifold). Look at its homology.

thanks! that clarifies @VladimirLysikov

sequentially compact intuitively in metric space combines with finite subcovers because when I have an infinite sequence of distinct elements then at least one subset within the finite subcover has to contain infinite elements of the sequence and this argument continiues recursively hence we end up with a cauchy sequence?

Now I have not one general way to generate Hausdorff non-regular extremally disconnected P-spaces from measurable cardinals, but two!

Turns out that if you add a non-open dense set to your topology, then the spaces stops being regular, but it still has all the other properties

So, given a Tychonoff extremally disconnected non-discrete P-space (existence of which is guaranteed by existence of measurable cardinals), all I had to do is find such dense non-open set (or equivalently, non-closed set with empty interior)

In the process I've proved that if $X$ is a discrete space of cardinality larger than the first measurable cardinal, then $\nu X\setminus X$ is not discrete

Another way of doing this, I didn't came up with it, David Gao did, you take strong ultrafilter topology on the set of all ultrafilters on a cardinal larger than the first measurable cardinal, and take its subspace consisting of all the $\sigma$-complete ultrafilters

strong ultrafilter topology is known to be extremally disconnected Hausdorff space which is not regular, this process takes all of its $P$-points, thus the subspace you take is a $P$-space, and since it's a dense subspace, it's extremally disconnected

the only non-trivial thing to check is that it's not regular, which can be done

so I have those two ways of creating such spaces, they're both just enrichments of topology on $\nu(\kappa)$ where $\kappa$ is larger than first measurable cardinal so perhaps they're not that different

then again, extremally disconnected spaces are famous for being almost always given in terms of ultrafilters

when where monoids mentioned